MAS1302 – Computational Probability and Statistics
Additional Exercises 3 (Not assessed)
These questions are designed for use in and around the third Problems Class. You do not need to hand in any
solutions, but the questions are good practice for the material covered being covered at the time.
1. Suppose U1 and U2 are two independent Uniform(0,1) random variables, and let Y=U1+U2. In lectures
we considered a geometrical proof of Pr(Y<1.5) = 7/8 = 0.875 (Example 3.2.1 Question 1).
(a) Without using a geometrical argument like the one used in lectures, state Pr(Y < 1).
Solution: Whatever the distribution of Y, it is clearly symmetrical about 1, so Pr(Y<1)=0.5.
(b) Back up your answer to part (a) with a geometrical proof.
Solution: Draw the unit square with axes for U1 and U2 , and the line U1+U2 = 1. The
required probability is the area of the lower left triangle, which is clearly 0.5.
2. Suppose U1 and U2 are two independent Uniform(0,1) random variables, and let Y=U1+U2. Evaluate
Pr(Y<0.25) justifying your answer with a geometrical proof.
Solution: Draw the unit square with axes for U1 and U2 , and the line U1+U2 = 0.25. The
required probability is the area of the lower left triangle, which is 0.5 x 0.25 x 0.25 = 1/32.
3. Suppose U1 and U2 are two independent Uniform(0,1) random variables.
(a) Without using a geometrical argument like the one used in lectures, state Pr(U1<U2).
Solution: By an obvious symmetry, Pr(U1<U2) = Pr(U1>U2), and hence each of these
quantities must be 0.5. [Note that we can exclude the possible outcome, U1=U2, because
although it is possible, it has zero probability, i.e. Pr(U1=U2) = 0.]
(b) Back up your answer to part (a) with a geometrical proof.
Solution: Draw the unit square with axes for U1 and U2 , and the line U1 = U2. The required
probability is the area of the upper left triangle, which is clearly equal to 0.5.
4. Suppose U1 and U2 are two independent Uniform(0,1) random variables.
(a) Evaluate Pr(U22 < 0.5).
Solution: U22 < 0.5 if and only if U2 < +√0.5 = 1/ √2 = 0.707. Now Pr(U2 < 0.707) = 0.707.
(b) Evaluate Pr(U22 < U1).
Solution: Draw the unit square with axes for U1 and U2, and the curve U22 = U1 i.e. U2 = √U1.
The required probability is the area under this curve, which is obtained as
1

0
x dx = 2/3.
5. Suppose U1 and U2 are two independent Uniform(0,1) random variables.
Evaluate Pr(U12 + U22 < 1).
Solution: Draw the unit square with axes for U1 and U2 , and the curve U12 + U22 = 1, which is a
quarter-circle (See Assignment 2, Part A, Task 1). The required probability is the area under
the curve, which is the area of a quarter-circle of radius 1, i.e. π/4. This can also be evaluated by
integration, and is a good test of your integration skills, while also providing a proof of the
formula for the area of a circle!
6. Suppose U1 , U2 and U3 are three independent Uniform(0,1) random variables.
Evaluate Pr(U12 + U22 + U32 < 1).
Solution: This is too hard to be considered useful for the exam, and is appropriate for people
with additional interest, so I am witholding the solution. It can be solved by analogy with the
solution given for Question 5 (here you need to know the volume of a sphere!). I am happy to
check individual solutions.
7. Suppose U1 and U2 are two independent Uniform(0,1) random variables, and let Y=U1+U2.
(a) Evaluate the p.d.f. of Y. (This is Example 3.2.1 Question 2).
(b) Using your answer to part (a), evaluate the c.d.f. of Y. (This is Example 3.2.1 Question 3).
Solution: This is too hard to be considered useful for the exam, and is appropriate for people
with additional interest, so I am witholding the solution. I am happy to check individual
solutions.
8. Suppose X1 ~ Bern(0.4);
X2 ~ Bern(0.6);
X3 ~ Bin(4, 0.4);
are independent random variables. Which of the following statements are true?
X4 ~ Bin(4,0.6)
(a) X1+X2 is a Binomial random variable;
Solution: False, because the success probabilities for X1 and X2 differ.
(b) X1+X3 is a Binomial random variable;
Solution: True, because the success probabilities for X1 and X3 are equal: X1+X3 ~ Bin(5, 0.4).
(c) X3+X4 is a Binomial random variable;
Solution: False, because the success probabilities for X3 and X4 differ.
(d) X3 can be expressed as the sum of four independent Bernoulli random variables.
Solution: True, X3 can be expressed as the sum of four independent Bern(0.4) random variables,
and in fact this is a good way of defining the Bin(4, 0.4) distribution!